Jake is planning a trip to china. he has made a list of cities he would like to visit, as well as the approximate amount of money he plans to spend in each as a result of travel, lodging, shopping, and so on. all costs are listed in renminbi (¥).

city
cost (¥)
tianjin
557
nanjing
681
zhengzhou
595
beijing
728
wuhan
449
chengdu
534

unfortunately, jake only has ¥2,920 available. what is the cheapest city that jake can remove from his travel plans and still stay under budget?
a.
tianjin
b.
wuhan
c.
nanjing
d.
beijing

Option C. Nanjing is the correct option.

Step-by-step explanation:

Jake is planning a trip to China. If we add expenses for all the cities then we get 557+681+595+728+449+534 = 3544, means Jake should have minimum 3544 yuan for the trip.

Unfortunately Jake has 2920 yuan available and he has to exclude one cheapest city from his trip.

Difference is 3544-2920 = 624 Therefore if he exclude Nanjing which has the expenditure of 681 yuan, his trip will remain under budget.

Option C. Nanjing is the answer.

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

i think it is c, not positive though

step-by-step explanation: